Hi Crack Dragon! Welcome to PF! Yes, it's a difference! The difference of a series {a_{n}} is the series {a_{n} - a_{n-1}} eg 0.5n(n+1) - 0.5(n-1)n = 0.5n((n+1) - (n-1)) = n
Hi, thanks for that, sorry if this was in the wrong area; wasn't too sure. That's pretty handy actually, by your post count you seem to be a busy person, but do you know where I could do any reading up on summations and the like?
Can you clarify, is the formula meant to hold for all n or for a fixed value of n? If it is for all n the f(n)=0 is the only solution in this case. If it is for a fixed n then there are infinite f(n) that will work.
Fixed n? I really am new to this, just in this format though if that's what you were wondering: http://www.wolframalpha.com/input/?i=Sum+f(n),+n=1+to+n, are you saying ∑0=1?
My mistake, I missread 1 as 0. But it doesn't change much. If the sum of f(k) from k=1 to n (btw, don't use n as both the thing you are summing over as well as the number of terms you are summing) for all n, then you must have f(1)=1 and all others are 0. To see this just set n=1 so f(1)=1 then set n=2 to get f(1)+f(2)=1 so 1+f(2)=1 which gives f(2)=0. Continue inductively to get the rest being 0.
[tex]\\ \sum f(n)\Delta n=1 \\ \\ \frac{\Delta\;\;}{\Delta n} \sum f(n)\Delta n = \frac{\Delta1}{\Delta n} \\ \\ f(n) = \left\{\begin{matrix} 1 & n=0\\ 0 & n\neq 0\\ \end{matrix}\right.[/tex]
The way you're using the summation symbol is wrong crack dragon. You need to sum over a "dummy" variable such as "k" in the example below. [tex] \sum_{k=1}^{n} f(k) = 1[/tex] Now we can again ask the question: Do you require this to be valid just for some specific positive "n" or is it to be true for all positive "n"? If it's for some specific "n" then there is a unique solution only for n=1, and an infinite number of solutions otherwise. If it's for all positive "n" then there is just one solution, that given above by jsmith.
##\Delta f=f_1-f_0## ##\frac{\Delta f}{\Delta x}=\frac{f_1-f_0}{x_1-x_0}## How in discrete calculus we not consider an infinitesimal interval but yes a discrete interval (unitary), so Δx=1, ie: ##\frac{\Delta f}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{f(x+1)-f(x)}{1}=f(x+1)-f(x)=\Delta f## The same logic for summation: ##\sum f(x)\Delta x=\sum f(x)\cdot 1=\sum _{x}f(x)## FTC for the discrete case: ##\sum_{x_0}^{x_1} f(x)\Delta x=\sum_{x=x_0}^{x_1}f(x)=F(x_1+1)-F(x_0)## inverse relationships: ##\sum \frac{\Delta f}{\Delta x}(x)\Delta x=f(x)## ##\frac{\Delta}{\Delta x} \sum f(x)\Delta x = f(x)## with infinitesimal limits... ##\int f(x)dx=\lim_{\Delta x \rightarrow 0} \sum f(x) \Delta x## ##\frac{df}{dx}=\lim_{\Delta x \rightarrow 0} \frac{\Delta f}{\Delta x}## computing... ##\frac{d}{dx}\left (\frac{1}{2}kx^2 \right )=kx## ##\frac{\Delta}{\Delta x}\left (\frac{1}{2}kx^2 \right )=kx+\frac{1}{2}k## ##\sum f(x)\Delta x=\frac{1}{6}kx^3-\frac{1}{4}kx^2+\frac{1}{12}kx + C## ##\int f(x)dx=\frac{1}{6}kx^3 + C## And you still have a Z transform that is the discrete analogue of Laplace transform. All this is the discrete calculus.
From the definition of discrete differentiation: ##\frac{\Delta f}{\Delta x} = f(x+1)-f(x)=f'(x)## If f(x)=1/2kx² so f'(x)=kx+1/2k Ironically or not, It's so that math works...
Now, my 2nd answer (considering the interval of summation): The "graphic" of question below is: ##F[n]=\sum _{1}^{n}f[m]\Delta m=1## \begin{matrix} n & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ F[n] & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\ F'[n] & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ \end{matrix} Thus we have ##F'[n]=\frac{\Delta F[n]}{\Delta n}=F[n+1]-F[n]=f[n]## ([...] the colchets says that the domain is discrete). Remember that: http://en.wikipedia.org/wiki/Kronecker_delta http://en.wikipedia.org/wiki/Dirac_delta_function ##\delta [n]:=\left\{\begin{matrix} 1 & n=0\\ 0 & n\neq 0 \end{matrix}\right.## So, f[n] in terms of discrete delta is: [tex]f[n]= \delta [n-1]=\left\{\begin{matrix} 1 & n=1\\ 0 & n\neq 1 \end{matrix}\right.[/tex] I found f(n) to you. Capiche?